Understanding Combinations
In combinatorics, the number of ways to choose $r$ objects from a set of $n$ distinct objects is denoted by $C(n, r)$ or $\binom{n}{r}$, defined as: $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$
The Powerful Ratio Property
A useful technique for solving equations involving consecutive combinations is the ratio method. Consider the ratio of two consecutive combinations: $$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n!}{r!(n-r)!} \cdot \frac{(r-1)!(n-r+1)!}{n!} = \frac{n-r+1}{r}$$
Step-by-Step Solution
Given:
- $\binom{n}{r-1} = 36$
- $\binom{n}{r} = 84$
- $\binom{n}{r+1} = 126$
Step 1: Solve for $r$
Divide (2) by (1): $$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{84}{36} = \frac{7}{3}$$ Using our ratio formula: $\frac{n-r+1}{r} = \frac{7}{3}$ $$3n - 3r + 3 = 7r \implies 3n = 10r - 3 \quad \text{(Equation A)}$$
Divide (3) by (2): $$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{126}{84} = \frac{3}{2}$$ Using the ratio formula (adjusted for $r+1$): $\frac{n-(r+1)+1}{r+1} = \frac{n-r}{r+1} = \frac{3}{2}$ $$2n - 2r = 3r + 3 \implies 2n = 5r + 3 \quad \text{(Equation B)}$$
Step 2: Solve the System
Multiply (B) by 2: $4n = 10r + 6 \implies 10r = 4n - 6$ Substitute into (A): $3n = (4n - 6) - 3$ $$3n = 4n - 9 \implies n = 9$$ Substitute $n=9$ into (B): $2(9) = 5r + 3 \implies 18 = 5r + 3 \implies 15 = 5r \implies r = 3$
Result: $n=9, r=3$.